Tại x = 2022 suy ra : 2023 = x + 1
Khi đó ta có : \(P\left(2022\right)\) \(=x^{x+1}-\left(x+1\right)x^x+\left(x+1\right)x^{x-1}-\left(x+1\right)x^{x-2}+...+\left(x+1\right)x-\left(x+1\right)\)
\(=x^{x+1}-x^{x+1}-x^x+x^x+x^{x-1}-x^{x-1}-x^{x-2}+...+x^2+x-x-1\)
\(=-1\)
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